A(2,-3,4), B(501), C(4,2,5) titik memabi AB sehingga AP:AB =2:3 panjang vektor AP?

AP : PB = m : n
maka
p = (mb + na)/(m + n)

A(2, -3, 4), B(5, 0, 1), C(4, 2, 5)
AP : AB = 2 : 3
PA : AB = -2 : 3
maka
a = (3p + (-2)b) / (3 + (-2))
a = (3p - 2b) / 1
a + 2b = 3p
p = (a + 2b)/3
p = ((2, -3, 4) + 2(5, 0, 1)) / 3
p = ((2, -3, 4) + (10, 0, 2)) / 3
p = (12, -3, 6)/3
p = (4, -1, 2)

AP = p - a
AP = (4, -1, 2) - (2, -3, 4)
AP = (2, 2, -2)
Panjang vektor AP
= |AP|
= √(2² + 2² + (-2)²)
= √(4 + 4 + 4)
= √12
= √4 . √3
= 2√3